The value of the integral ∫ limits0(π/2) 60 ( sin (6 x)/ sin x) d x is equal to

Question

The value of the integral ∫ limits0(π/2) 60 ( sin (6 x)/ sin x) d x is equal to (A)  (B)  (C)  (D)

Tardigrade Admin · Accepted Answer

I=60 ∫ limits0π / 2(( sin 6 x- sin 4 x/ sin x)+( sin 4 x- sin 2 x/ sin x)+( sin 2 x/ sin x)) d x I=60 ∫ limits0π / 2(2 cos 5 x+2 cos 3 x+2 cos x) d x I=.60((2/5) sin 5 x+(2/3) sin 3 x+2 sin x)|0 π / 2=104

Q. The value of the integral $0 \int \frac{π}{2} 60 \frac{s i n ( 6 x )}{s i n x} d x$ is equal to

$I = 60 0 \int π /2 (\frac{s i n 6 x - s i n 4 x}{s i n x} + \frac{s i n 4 x - s i n 2 x}{s i n x} + \frac{s i n 2 x}{s i n x}) d x$
$I = 60 0 \int π /2 (2 cos 5 x + 2 cos 3 x + 2 cos x) d x$
$I = 60 (\frac{2}{5} sin 5 x + \frac{2}{3} sin 3 x + 2 sin x) ∣ ∣_{0}^{π /2} = 104$

Q. The value of the integral 0∫2π​​60sinxsin(6x)​dx is equal to

I=600∫π/2​(sinxsin6x−sin4x​+sinxsin4x−sin2x​+sinxsin2x​)dx I=600∫π/2​(2cos5x+2cos3x+2cosx)dx I=60(52​sin5x+32​sin3x+2sinx)∣∣​0π/2​=104

Q. The value of the integral $0 \int \frac{π}{2} 60 \frac{s i n ( 6 x )}{s i n x} d x$ is equal to

$I = 60 0 \int π /2 (\frac{s i n 6 x - s i n 4 x}{s i n x} + \frac{s i n 4 x - s i n 2 x}{s i n x} + \frac{s i n 2 x}{s i n x}) d x$
$I = 60 0 \int π /2 (2 cos 5 x + 2 cos 3 x + 2 cos x) d x$
$I = 60 (\frac{2}{5} sin 5 x + \frac{2}{3} sin 3 x + 2 sin x) ∣ ∣_{0}^{π /2} = 104$